Question

In a hydrogen atom, the electron makes $6.6\mathrm{x}{10}^{15}$ revolutions per second around the nucleus in an orbit of radius $0.5\mathrm{x}{10}^{-10}\mathrm{m}$. It is equivalent to a current nearly:

• A. $1$ Ampere
• B. $1$ milliAmpere
• C. $1$ microAmpere
• D. $1.6x{10}^{-19}$ Ampere

Answer 1

The correct option is B

$1$ milliAmpere

Step 1: Given Data

Given the frequency of the electron $\mathrm{f}=$$6.6\mathrm{x}{10}^{15}$ revolutions per second

The radius of the orbit $\mathrm{r}=$$0.5\mathrm{x}{10}^{-10}\mathrm{m}$

Step 2: Calculate the Current

If $\mathrm{i}$ is current, $\mathrm{q}$ is the charge of an electron, and $\mathrm{t}$ is the time taken, then we know that,

$\mathrm{i}=\frac{\mathrm{q}}{\mathrm{t}}$

We know that charge of an electron is $\mathrm{q}=1.6\times {10}^{-19}\mathrm{C}$

$\therefore$Time period $\mathrm{t}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$

Where $\mathrm{v}$ is the velocity.

Therefore, frequency $\mathrm{f}=\frac{1}{\mathrm{t}}=\frac{\mathrm{v}}{2\mathrm{\pi r}}$

Hence, $\mathrm{i}=\mathrm{q}\times \mathrm{f}$

Upon substituting the values we get,

$\mathrm{i}=1.6\times {10}^{-19}\times 6.6\times {10}^{15}=1.05mA$

$\approx 1$milliAmpere

Hence, the correct answer is option (B).